1031 Hello World for U (20 point(s))

本文链接： https://blog.csdn.net/huaxuewan/article/details/101015540

Given any string of N (≥ 5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as:

h !
e d
l l
lowor

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n₁ characters, then left to right along the bottom line with n₂ characters, and finally bottom-up along the vertical line with n₃ characters. And more, we would like U to be as squared as possible – that is, it must be satisfied that n₁ = n₃ = max { k | k ≤ n₂ for all 3 ≤ n₂ ≤ N } with n₁ + n₂ + n₃ − 2 = N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

Sample Output:

h !
e d
l l
lowor

题目大意：

输入的字符串，按照 U 字型输出；

n₁ 和 n₃ 表示两条竖线上字符的总个数，n₂ 表示底部横线上字符的总个数，并要求：

n₁ == n₃
n₂ >= n₁
n₁ 取尽可能大的值

设计思路：

难点在于确定 n₁、n₂ 的值；

因为 U 分为 n₁、n₂、n₃ 三条边，且重叠了两个字符，所以设：

n = 原字符串长度 + 2 = n₁ * 2 + n₂；
- 若 n % 3 == 0，则 n₁ == n₂ == n₃；
- 若 n % 3 > 0，并且需要 n₂ >= n₁，那么把多出的余数给 n₂ 就行了；
所以 n₁ = n / 3，n₂ = n / 3 + n % 3；

确定了三条边的长度，利用一个头指针和一个尾指针，按行输出原字符串即可

编译器：C (gcc)

#include <stdio.h>
#include <string.h>

int main(void)
{
        char str[81], *p, *q;
        int n1, n2, count;
        int i, j;

        scanf("%s", str);

        count = strlen(str) + 2;
        p = str;
        q = str + (count - 2 - 1);
        n1 = count / 3;
        n2 = count / 3 + count % 3;
        for (i = 0; i < n1 - 1; i++) {
                printf("%c", *p);
                p++;
                for (j = 1; j < n2 - 1; j++)
                        printf(" ");
                printf("%c\n", *q);
                q--;
        }
        for (p; p <= q; p++)
                printf("%c", *p);

        return 0;
}